# Question regarding solving multivariable function continuity problems

$$f(x,y)=\begin{Bmatrix} \frac{x+y}{x^2+y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

Let's say I have this problem. I use $$y = mx$$, solving this I get $$\frac{1+m}{x(1+m^2)}$$. Calculating the limit I get $$\frac{1+m}{0}$$. Could this prove that the limit doesn't exist?

Some follow-up questions:

• When can I use th $$y=mx$$ method, if I can?
• What is the rule of paths? I guess I can't just plug in $$(1,2)$$ or $$(2,3)$$, but can I plug in let's say $$(\frac{1}{n}, \frac{1}{n})$$? If yes, what should I look for when plugging in values? The 'n' has to go, but how?
• If I need to prove the continuity of a function, what is the simplest way to prove it? I heard about the Squeeze Theorem, but does that work outside of sin, cos functions? At seminars we used $$a^2 + b^2 >= 2ab$$, but I don't find that at all clear.

EDIT:

$$f(x,y)=\begin{Bmatrix} \frac{x^2y3}{x^2+y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

I have this function. Using $$a^2 + b^2 >= 2ab$$:

$$x^2 + y^2 >= |x^2| + |y^2| >= 2|x||y|$$

$$|f(x,y)| = \frac{x^2 |y^3|}{x^2 + y^2} = \frac{x^2 |y^3|}{2|x||y|} = \frac{x y^2}{2}$$ ---> this tending to 0 would prove continuity? if yes, does this method work all the time?

EDIT 2:

$$f(x,y)=\begin{Bmatrix} \frac{3x^2y}{2x^2+5y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

$$|f(x,y)| = \frac{3x^2|y|}{2x^2+5y^2} <= \frac{3x^2|y|}{2\sqrt{10}|x||y|} = \frac{3}{2\sqrt{10}}*x$$

$$(a^2 + b^2 >= 2ab) ==> 2x^2 + 5y^2 >= (|x|\sqrt{2})^2 + (|y|\sqrt{5})^2 >= 2\sqrt{10}*x$$

$$f(x,y) <= \frac{3}{2\sqrt{10}}*x$$(1)

$$x --> 0, y --> 0$$ (2)

(1)(2) --> f(x,y) --> 0 = f(0, 0).

1. Given a function $$f$$ on $$\mathbb{R}^2$$, it does not make sense to say "the limit" of $$f$$: you should indicate the limit at what point.

2. I don't know how you get $$\frac{1+m}{0}$$. But restricting on $$y=x$$, you have $$f(x,y)=\frac{1}{x}$$, which tells you that the limit $$\displaystyle \lim_{(x,y)\to(0,0)}f(x,y)$$ does not exist.

3. In real analysis, there is usually no one-fit-all "method". What you often have is a certain set of "strategies". You usually try the path $$y=mx$$ for some fixed $$m$$ when studying functions of the form $$f(x,y)=\frac{P}{Q}$$ where $$P$$ and $$Q$$ are polynomials in $$x$$ and $$y$$.

4. Again, there is no "rule". You need to work on an ad hoc basis.

5. "If I need to prove the continuity of a function, what is the simplest way to prove it?" There is no such thing. Sometimes, you would need to work from the first principle.

6. In your edited question, the estimate $$|\frac{x^2y^3}{x^2+y^2}|\le \frac{|x^2|y^3|}{2|x||y|}$$ shows that $$f$$ is continuous at $$(0,0)$$.

• thank you, I know my terminology is a tad off, sorry for that. Yes, I meant continuous at (0, 0), and looking for limits at (0, 0). Jan 12, 2021 at 14:28
• @samivagyok: no problem! I was just trying to make it precise. :-)
– user9464
Jan 12, 2021 at 14:29
• @samivagyok Anything helpful for you here?
– user9464
Jan 12, 2021 at 14:30
• Yes, it was. So if I need to find continuity in a point let's say (0, 0), can I use y = mx, y = x etc.? And at the edited segment was the calculation correct? Jan 12, 2021 at 15:06
• @samivagyok: not quite, in general, you basically work with the definition.// for this particular problem, you work with $y=mx$ for some $m$; for your edited question, no// it should be useful to see more examples.
– user9464
Jan 12, 2021 at 15:07